Given two arrays A[] and B[] consisting of N integers where A[i] represents the initial position of the ith plane and B[i] is the speed at which the plane is landing, the task is to print the number of the plane that can be stopped from landing by shooting an aircraft at every second.
Examples:
Input: A[] = {1, 3, 5, 4, 8}, B[] = {1, 2, 2, 1, 2}
Output: 4
Explanation:
- At second 1, shoot the plane at index 0, the positions of the planes are now A[] = {_, 1, 3, 3, 6}.
- At second 2, shoot the plane at index 1, the positions of the planes are now A[] = {_, _, 1, 2, 4}.
- At second 3, shoot the plane at index 2, the positions of the planes are now A[] = {_, _, _, 1, 2}.
- At second 4, shoot the plane at index 4 or 5, the positions of the planes are now A[] = {_, _, _, _, _}.
Therefore, a total of 4 planes can be stopped from landing.
Input: A[] = {2, 8, 2, 3, 1}, B[] = {1, 4, 2, 2, 2}
Output: 2
Approach: The given problem can be solved based on the following observations:
- It can be observed that the count of planes that can be stopped is the set of distinct times needed to land for each plane, as at the same time, only one plane can be destroyed.
- The time needed for a plane to land is A[i] / B[i] can be calculated using the formula: Time = Distance / Velocity
Follow the steps below to solve the problem:
- Initialize a Hash-Set, say S that stores the times needed for plane landing.
-
Iterate over a range [0, N] using the variable i and perform the following tasks:
- Initialize a variable, say t as 1 if the value of A[i]%B[i] is greater than 0. Otherwise, update the variable t as 0.
- Add the value of A[i]/B[i] into the variable, say t.
- Add the value of t into the HashSet S.
- After performing the above steps, return the size of the HashSet as the resultant answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;
// Function to find maximum number// of planes that can be stopped// from landingint maxPlanes(int A[], int B[], int N)
{ // Stores the times needed for
// landing for each plane
set<int> St;
// Iterate over the arrays
for (int i = 0; i < N; i++) {
// Stores the time needed for
// landing of current plane
int t = (A[i] % B[i] > 0) ? 1 : 0;
// Update the value of t
t += (A[i] / B[i]) + t;
// Append the t in set St
St.insert(t);
}
// Return the answer
return St.size();
} // Driver Code
int main() {
int A[] = { 1, 3, 5, 4, 8 };
int B[] = { 1, 2, 2, 1, 2 };
int N = sizeof(A)/sizeof(A[0]);
cout << maxPlanes(A, B, N);
return 0;
}// This code is contributed by Dharanendra L V. |
Java
// Java program for the above approachimport java.util.*;
class GFG {
// Function to find maximum number
// of planes that can be stopped
// from landing
static int maxPlanes(int[] A, int[] B)
{
// Stores the times needed for
// landing for each plane
Set<Integer> St = new HashSet<>();
// Iterate over the arrays
for (int i = 0; i < A.length; i++) {
// Stores the time needed for
// landing of current plane
int t = (A[i] % B[i] > 0) ? 1 : 0;
// Update the value of t
t += (A[i] / B[i]) + t;
// Append the t in set St
St.add(t);
}
// Return the answer
return St.size();
}
// Driver Code
public static void main(String[] args)
{
int[] A = { 1, 3, 5, 4, 8 };
int[] B = { 1, 2, 2, 1, 2 };
System.out.println(
maxPlanes(A, B));
}
} |
Python3
# Python program for the above approach# Function to find maximum number# of planes that can be stopped# from landingdef maxPlanes(A, B, N):
# Stores the times needed for
# landing for each plane
St = set()
# Iterate over the arrays
for i in range(N):
# Stores the time needed for
# landing of current plane
t = 1 if (A[i] % B[i] > 0) else 0
# Update the value of t
t += (A[i] // B[i]) + t
# Append the t in set St
St.add(t)
# Return the answer
return len(St)
# Driver CodeA = [ 1, 3, 5, 4, 8 ]
B = [ 1, 2, 2, 1, 2 ]
N = len(A)
print(maxPlanes(A, B, N))
# This code is contributed by shivanisinghss2110 |
C#
// C# program for the above approachusing System;
using System.Collections.Generic;
public class GFG
{ // Function to find maximum number
// of planes that can be stopped
// from landing
static int maxPlanes(int[] A, int[] B)
{
// Stores the times needed for
// landing for each plane
HashSet<int> St = new HashSet<int>();
// Iterate over the arrays
for (int i = 0; i < A.Length; i++) {
// Stores the time needed for
// landing of current plane
int t = (A[i] % B[i] > 0) ? 1 : 0;
// Update the value of t
t += (A[i] / B[i]) + t;
// Append the t in set St
St.Add(t);
}
// Return the answer
return St.Count;
}
// Driver code
static public void Main (){
int[] A = { 1, 3, 5, 4, 8 };
int[] B = { 1, 2, 2, 1, 2 };
Console.WriteLine(
maxPlanes(A, B));
}
}// This code is contributed by Potta Lokesh |
Javascript
<script> // Function to find maximum number
// of planes that can be stopped
// from landing
function maxPlanes(A, B)
{
// Stores the times needed for
// landing for each plane
let St = new Set();
// Iterate over the arrays
for (let i = 0; i < A.length; i++) {
// Stores the time needed for
// landing of current plane
let t = (A[i] % B[i] > 0) ? 1 : 0;
// Update the value of t
t += Math.floor(A[i] / B[i]) + t;
// Append the t in set St
St.add(t);
}
// Return the answer
return St.size;
}
// Driver Code let A = [ 1, 3, 5, 4, 8 ];
let B = [ 1, 2, 2, 1, 2 ];
document.write(
maxPlanes(A, B));
// This code is contributed by sanjoy_62.
</script> |
Output:
3
Time Complexity: O(N*log(N))
Auxiliary Space: O(N)